Created
January 5, 2015 12:57
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distance from point to line segment…
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from numpy import arccos, array, dot, pi | |
from numpy.linalg import det, norm | |
def distance(A, B, P): | |
""" segment line AB, point P, where each one is an array([x, y]) """ | |
if all(A == P) or all(B == P): | |
return 0 | |
if arccos(dot((P - A) / norm(P - A), (B - A) / norm(B - A))) > pi / 2: | |
return norm(P - A) | |
if arccos(dot((P - B) / norm(P - B), (A - B) / norm(A - B))) > pi / 2: | |
return norm(P - B) | |
return abs(dot(A - B, P[::-1]) + det([A, B])) / norm(A - B) |
The last return on line 12 was wrong. This works instead:
from numpy import arccos, array, dot, pi, cross from numpy.linalg import det, norm # from: https://gist.github.com/nim65s/5e9902cd67f094ce65b0 def distance_numpy(A, B, P): """ segment line AB, point P, where each one is an array([x, y]) """ if all(A == P) or all(B == P): return 0 if arccos(dot((P - A) / norm(P - A), (B - A) / norm(B - A))) > pi / 2: return norm(P - A) if arccos(dot((P - B) / norm(P - B), (A - B) / norm(A - B))) > pi / 2: return norm(P - B) return norm(cross(A-B, A-P))/norm(B-A)
You did not handle the case where A equals B, which throws a true_divide
. Something like this should be added:
if all(A == B): return dist(A, P)
I would also add a np.clip
within arccos
such that: arccos(np.clip(angle, -1, 1))
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The last return on line 12 was wrong. This works instead: